# Sum of divisors of perfect square

Let $$c,d$$ be natural numbers of same parity (both odd or both even) and $$\sigma$$ be sum of divisors function. Is it known whether or under what conditions $$\sigma (c^{2})$$=$$\sigma (d^{2})$$ ? I am guessing that it can happen but unsure if certain things must hold

• Have you seen this? – Don Thousand Dec 24 '18 at 19:19
• Thank you,I don't know if that helps me – argamon Dec 24 '18 at 19:21
• @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$ – AgentS Dec 24 '18 at 20:08
• I bet you had tried that, not that interesting.. – AgentS Dec 24 '18 at 20:12
• Haha, it's just more useful if I know in general but I greatly appreciate your help – argamon Dec 24 '18 at 20:13

It looks to me like $$\sigma(627^2)=\sigma(749^2)$$.

Feel free to check my work. (Or Python's work!)

And here's an even-even pair: $$\sigma(740^2)=\sigma(878^2)$$.

As pointed out in the comments, starting from the simple example $$\sigma (4^2)=\sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $$10$$. Thus, $$\sigma(12^2)=(1+2+2^2+2^3+2^4)\times (1+3+3^2)=31\times 13=403$$ $$\sigma(15^2)=(1+3+9)\times (1+5+25)=13\times 31=403$$

and so on.

• That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even – argamon Dec 24 '18 at 19:45
• I suggest doing a search. I also found $\sigma(76^2)=\sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that. – lulu Dec 24 '18 at 19:48
• In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair. – Ross Millikan Dec 24 '18 at 20:22
• You have $\sigma(4^2)=\sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$ – Ross Millikan Dec 24 '18 at 20:44
• @RossMillikan Yes, both my examples are simple consequences of \$\sigma(4^2)=\sigma (5^2). I'll edit to point that out, – lulu Dec 24 '18 at 20:52